Question: KC has a piece of gum stuck to her bike tire, which applies a forward torque (i.e. a force to roll forwards) on the tire's movement. The torque varies in a periodic way that can be modeled approximately by a trigonometric function. When the gum is on the front of the tire, its weight is pulling forwards with a maximum torque of $0.01\text{ Nm}$ (Newton metre, the SI unit for torque), and when it's on the back of the tire, it's pulling backwards with a minimum torque of $-0.01\text{ Nm}$. The maximum torque is reached once in every rotation, which is every $1.2$ meters. The first time it reaches its maximum torque is $0.3$ meters into the race. Find the formula of the trigonometric function that models the torque $\tau$ the gum applies on the tire $d$ meters into the race. Define the function using radians. $ \tau(d) = $ What is the torque when KC has ridden $4$ meters? Round your answer, if necessary, to four decimal places. $ $
Answer: Let's start by writing a formula for the torque $x$ meters after it reaches its maximum. Both sine and cosine can be used to model periodic contexts. We can decide which is better fitting by considering the $y$ -intercept. The sine function intercepts the $y$ -axis at its midline, and the cosine function intercepts the $y$ -axis at its peak. We know the torque reaches its highest point at $x = 0$, so let's use a cosine function. The amplitude of the torque function is $\dfrac{0.01 - (-0.01)}{2} = 0.01$ $\text{cm}$. The period is $1.2$ meters, since a cosine function reaches its peak once in every period. The midline is the average of the highest and lowest values, or $\dfrac{(0.01) + (-0.01)}{2} = 0$ Since the ordinary cosine function $f(x) = \cos x$ has period $2\pi$, midline $y = 0$, and amplitude $1$, we need to stretch it horizontally by a factor of ${\dfrac{1.2}{2\pi}}$ and stretch it vertically by a factor of ${0.01}$. $ \tau(x) = {0.01}\cos\left({\dfrac{2\pi}{1.2}}x\right)$ Since the torque reaches its maximum after $0.3$ meters, when KC is $d$ meters into the race she's traveled $d - 0.3$ meters from the maximum torque. So $x = d - 0.3$ : $ \tau(d) = {0.01}\cos\left({\dfrac{2\pi}{1.2}}(d-0.3)\right)$ After $4$ meters, the torque from the piece of gum is $\begin{aligned} \tau(4) &= 0.01\cos\left(\dfrac{2\pi}{1.2}(4-0.3)\right)\\ &\approx 0.0087 \end{aligned}$ A correct formula for $\tau(d)$ is: $ \tau(d) = 0.01\cos\left(\dfrac{2\pi}{1.2}(d-0.3)\right)$ The torque when KC has ridden $4$ meters is: $ 0.0087\text{ Nm}$